3.17 \(\int (a+a \cos (c+d x))^2 \, dx\)

Optimal. Leaf size=45 \[ \frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{3 a^2 x}{2} \]

[Out]

(3*a^2*x)/2 + (2*a^2*Sin[c + d*x])/d + (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.0140222, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2644} \[ \frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{3 a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^2,x]

[Out]

(3*a^2*x)/2 + (2*a^2*Sin[c + d*x])/d + (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^2 \, dx &=\frac{3 a^2 x}{2}+\frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0423232, size = 34, normalized size = 0.76 \[ \frac{a^2 (6 (c+d x)+8 \sin (c+d x)+\sin (2 (c+d x)))}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^2,x]

[Out]

(a^2*(6*(c + d*x) + 8*Sin[c + d*x] + Sin[2*(c + d*x)]))/(4*d)

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Maple [A]  time = 0.039, size = 52, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,{a}^{2}\sin \left ( dx+c \right ) +{a}^{2} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^2,x)

[Out]

1/d*(a^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+2*a^2*sin(d*x+c)+a^2*(d*x+c))

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Maxima [A]  time = 1.1332, size = 61, normalized size = 1.36 \begin{align*} a^{2} x + \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{4 \, d} + \frac{2 \, a^{2} \sin \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

a^2*x + 1/4*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2/d + 2*a^2*sin(d*x + c)/d

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Fricas [A]  time = 1.60468, size = 82, normalized size = 1.82 \begin{align*} \frac{3 \, a^{2} d x +{\left (a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(3*a^2*d*x + (a^2*cos(d*x + c) + 4*a^2)*sin(d*x + c))/d

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Sympy [A]  time = 0.252579, size = 78, normalized size = 1.73 \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + a^{2} x + \frac{a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 a^{2} \sin{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + a\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**2,x)

[Out]

Piecewise((a**2*x*sin(c + d*x)**2/2 + a**2*x*cos(c + d*x)**2/2 + a**2*x + a**2*sin(c + d*x)*cos(c + d*x)/(2*d)
+ 2*a**2*sin(c + d*x)/d, Ne(d, 0)), (x*(a*cos(c) + a)**2, True))

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Giac [A]  time = 1.36878, size = 51, normalized size = 1.13 \begin{align*} \frac{3}{2} \, a^{2} x + \frac{a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{2 \, a^{2} \sin \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

3/2*a^2*x + 1/4*a^2*sin(2*d*x + 2*c)/d + 2*a^2*sin(d*x + c)/d